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# Case Problem Par, Inc. Par, Inc., is a major manufacturer of golf equipment. Management believes that Pars market share could be increased with

## Case Problem Par, Inc.

Par, Inc., is a major manufacturer of golf equipment. Management believes that Par’s market share could be increased with the introduction of a cut-resistant, longer-lasting golf ball. Therefore, the research group at Par has been investigating a new golf ball coating designed to resist cuts and provide a more durable ball. The tests with the coating have been promising.

One of the researchers voiced concern about the effect of the new coating on driving distances. Par would like the new cut-resistant ball to offer driving distances comparable to those of the current-model golf ball. To compare the driving distances for the two balls, 40 balls of both the new and current models were subjected to distance tests. The testing was performed with a mechanical hitting machine so that any difference between the mean distances for the two models could be attributed to a difference in the two models. Theresults of the tests, with distances measured to the nearest yard, follow. These data are available in the file Golf.

Do:

In a managerial report, use the methods of hypothesis testing to

• Provide descriptive statistical summaries of the data for each model, written in complete sentences.
• Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.
• Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test? What is your recommendation for Par, Inc.?
• Discuss whether you see a need for larger sample sizes and more testing with the golf balls.

Discuss

Post by classmate 1

Hello everyone,

Here are my findings and managerial report after using the methods of hypothesis testing:

After reviewing the data for both current and new golf ball driving distance, a t-Test was performed to determine if the current golf balls travel a longer distance than the new ones in a sample size of 40. Assuming that the data is normally distributed, the t-Test displayed a sample mean for the current golf ball of 270.28 and a sample mean for the new golf ball of 267.5. The difference between the means is not significant. The variance for the current golf ball is 76.61, and the variance for the new golf ball is 97.95. The p-value is 0.094. The standard deviation for the current golf ball is 8.75, and the standard deviation for the new golf ball is 9.9. The “t Critical one-tail” is 1.66, and the “t Critical two-tail” is 1.99.

The hypothesis test was needed to determine if there is enough evidence to support that the current golf balls travel a greater distance than the new ones proposed for Par, Inc (null hypothesis) or if, in fact, they are the same (alternative hypothesis).

The p-value is a random variable derived from the distribution and is a measure of evidence against the null hypothesis (Hung et al., 1997). In this case, the p-value is 0.094, which is less than 0.05, indicates strong evidence against the null hypothesis as there is less than a 5% probability that the null would be correct; hence, the p-value is greater than the significance level, and one should fail to reject the null hypothesis (McLeod, 2019). There appears no evidence to support that the current golf balls travel a more significant distance than the new ones. Given the existing data, there is no evidence to support that the alternative hypothesis is correct, as one would need more data.

Based on my hypothesis testing conclusion, my recommendations for Par, Inc would be to utilize a more extensive-sized data sampling for current and new golf balls. An increased sample size would decrease the standard error of associated sampling distributions (Anderson et al., 2021).

References

Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., Cochran, J. J., Fry, M. J., & Ohlmann. J. W. (2021). Essentials of modern business statistics with Microsoft® Excel® (8th ed.). Cengage Learning

H. M. James Hung, O’Neill, R. T., Bauer, P., & Kohne, K. (1997). The Behavior of the P-Value When the Alternative Hypothesis is True. Biometrics53(1), 11–22. https://doi.org/10.2307/2533093 (Links to an external site.)

McLeod, S. (2019). What a p-value tells you about statistical significance. P. https://www.simplypsychology.org/p-value.html

Post by classmate 2

Wk2 Discussion

• Provide descriptive statistical summaries of the data for each model, written in complete sentences.
• From the data that was given we compared two different types of golf balls to see if there is a difference in the current ones being used and the new durable cut resistant ones. The data compares 40 different golf ball drives with each test being for each of the golf balls. When comparing the statistical data given, The current golf ball had an average distance of 270.28 versus the new golf balls having an average distance of 267.48 which is a drop of 1.04% for averages. For highest drive distance per choice was the same at 289, while the low had a difference of 255 for the old ones versus 250 for the newer gold balls, this also had a decrease of 1.96%. The median between the two golf balls was similar to the average at 270 for the current and 265 for the newer ones. On average the change between the current golf balls to the new, more durable golf balls was each of the drives lost an average of 2.8. The p-value is 0.094, T-Critical one-tail is 1.66, and the T-Critical two-tail is 1.99.
• Formulate and present the rationale for a hypothesis test that Par could use to compare the driving distances of the current and new golf balls. Clearly state the null and alternative hypothesis.
• Null hypothesis: Current golf balls that Par uses travel a greater distance than the more durable, cut resistant golf balls.
• Alternate hypothesis: New golf balls travel the same distance as current ones.
• Analyze the data to provide the hypothesis testing conclusion. What is the p-value for your test?
• With the p-value (0.094) being more than the (0.05) 5% probability making the null hypothesis not proven and would lean more towards the alternate hypothesis. This means that the data is not showing a lower than 5% accuracy making the null hypothesis true.
• What is your recommendation for Par, Inc.? Discuss whether you see a need for larger sample sizes and more testing with the golf balls.
• With a test of this size showing almost identical results with little differences, a large scale would give a more accurate account. Based on there really only having such a small change in distance, many variable could have been at play to skew these results. If all the hits were the same order, golfers fatigue, and wind speed being some of these variables. If the scale was out of 100 than the results would supply more accurate result since it would give them an out of 100% answer they would need to confirm this theory. They would have to test the on almost identical condition days with the golfer hitting the ball in one order the first time and opposite order the second time. This would eliminate a lot of the “what if” variables that affect the overall numbers on this (mythbusters logic).
• Reference:
• Anderson, D. R., Sweeney, D. J., Williams, T. A., Camm, J. D., Cochran, J. J., Fry, M. J., & Ohlmann. J. W. (2021). Essentials of modern business statistics with Microsoft® Excel® (8th ed.). Cengage Learning